Verify by calculating that :
Question:

Verify by calculating that :

(a) 5 moles of $\mathrm{CO}_{2}$ and 5 moles of $\mathrm{H}_{2} \mathrm{O}$ do not have the same mass.

(b) $240 \mathrm{~g}$ of calcium and $240 \mathrm{~g}$ magnesium elements have mole ratio of $3: 5$.

Solution:

(a) Mass of 5 moles of $\mathrm{CO}_{2}$

Mass of 1 mole of $\mathrm{CO}_{2}=12+2 \times 1=44 \mathrm{~g} \mathrm{~mol}^{-1}$.

Mass of 5 moles of $\mathrm{CO}_{2}=(5 \mathrm{~mol}) \times\left(44 \mathrm{~g} \mathrm{~mol}^{-1}\right)=220 \mathrm{~g}$

Mass of 5 moles of $\mathrm{H}_{2} \mathrm{O}$

Mass of 1 mole of $\mathrm{H}_{2} \mathrm{O}=2 \times 1+16=18 \mathrm{~g} \mathrm{~mol}^{-1}$.

Mass of 5 moles of $\mathrm{H}_{2} \mathrm{O}=(5 \mathrm{~mol}) \times\left(18 \mathrm{~g} \mathrm{~mol}^{-1}\right)=90 \mathrm{~g}$

The two values are not equal.

$(b)$ 

Mass of calcium $(\mathrm{Ca})=240 \mathrm{~g}$

Molar atomic mass of calcium $(\mathrm{Ca})=40 \mathrm{~g} \mathrm{~mol}^{-1}$.

No. of moles of calcium $(\mathrm{Ca})=\frac{(240 \mathrm{~g})}{\left(40 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=6 \mathrm{~mol}$

Mass of magnesium $(\mathrm{Mg})=240 \mathrm{~g}$

Molar atomic mass of magnesium $(\mathrm{Mg})=24 \mathrm{~g} \mathrm{~mol}^{-1}$

No. of moles of magnesium $(\mathrm{Mg})=\frac{(240 \mathrm{~g})}{\left(24 \mathrm{~g} \mathrm{~mol}^{-1}\right)}=10 \mathrm{~mol}$

Ratio may be given as $6 \mathrm{~mol}: 10 \mathrm{~mol}$ or $3: 5$.

 

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